text:
How did Wilbur Wright create and use the formulas he mentions in his letter to Octave Chanute?
In a letter written by Wilbur Wright and addressed to Octave Chanute, dated 1903/06/18, I have found two things that I do not quite understand.

Our engine develops at the brake 15.6 horse power and we are convinced that this is very close to what we will be able to reach as a maximum. It furnishes in foot pounds at the brake, energy equivalent to 23 percent of the heat units
  contained in the fuel, so its efficiency is some twenty percent higher than the usual efficiency of gasoline engines which ranges from 14 to 18 percent of the total power contained in the fuel.

How did the two brothers make such a revolutionary improvement?
Modern gasoline engines have a maximum thermal efficiency of about 25% to 30%. The Wright brother's motor had an efficiency of 23% in 1903.

The Dumont motor may develop more than 10 horsepower but it is with me a case of seeing before believing. The screw of the airship was 4 meters in diameter and had a pitch of 4 meters, and the engine at 1200 revolutions (which is the speed of maximum power of motors of this size) drove it 200 turns per minute and gave a thrust of 154 to 165 lbs. Now if you multiply the thrust, by the pitch divided by the circumference of the screw, and multiply this by the travel per minute of the center of pressure (which is about 5/6 of the circumference of the screw) you will find that the result is about 10 horse power.”

It is evident that the Wrights used the relation:
Power = Thrust x (5/6) x pitch x (RPM/60 sec)
Where does this formula come from?
What exactly is that "center of pressure (which is about 5/6 of the circumference of the screw)"?

I do not have first-hand knowledge of this formula, so what I write below is speculation. But the Wrights used sound physical reasoning, and I will try do do the same.
Power in physics is the product of force and speed. The formula does indeed contain a force (thrust) and a speed (pitch times rotation speed) if the terms used by Wilbur are interpreted in a certain way. Pitch today is an angle, but I would think Wilbur meant with pitch what is termed distance D in the picture below.

Propeller advance ratio (picture source)
This is consistent with his earlier sentence where he gives a length for the term pitch. The term pitch * RPM/60 is actually the flight speed in m/s.
Now to the term 5/6: I would expect that Wilbur calculated the pitch from the incidence of his propeller. Since the propeller blade needs some angle of attack to produce thrust, the actual distance traveled through air is a little smaller than what Wilbur got. Therefore, I think the term 5/6 accounts for that slippage and had been determined empirically.
Now for the revolutionary increase in efficiency: I do not know what engines Wilbur used for comparison, but the low number he cites suggests that he used exceptionally poor specimens .

Peter Kämpf, I added more explanations to the text I posted above. (1) W. Wright estimated the engine used by Santos Dumont delivered about 10 hp. The calculations are not about a motor built in Dayton. It appears that, without realizing, W. Wright forced Santos' airship to travel at (5/6) x pitch x (RPM/60 sec) = 24.85 mph. (2) The Wright brother's motor had an efficiency of 23% in 1903! Modern gasoline motors efficiency ranges from 25 to 30 hp.


text:
What was the exact propeller geometry of the 1903 Wright Flyer? A modern replica failed the test T = D in a NASA wind tunnel
Full-Scale 1903 Wright Flyer failed the test Thrust = Drag in a NASA wind tunnel
I have found a presentation about an extensive set of tests done in 1999 on a full scale replica of Flyer I 1903. The plane did not behave well and was not able to produce enough thrust to overcome the drag.
"In the wind tunnel, the reproduced propeller fell short of T=D @ 28 mph test condition when operating at the maximum permitted speed of 340 RPM. These data are being analyzed"

$C_L=C_L(C_D)$ Flyer I replica
Source: Full-Scale 1903 Wright Flyer Wind Tunnel Test Results From the NASA Ames Research Center
Unfortunately, the presentation, also quite explicit, does not give the values of the Thrust, Drag or Drag - Thrust at 28 mph. I calculated the Drag from $C_L=C_L(C_D)$, Lift = Weight (the mass of the airplane is known) and and the surface of the wings. Using a simulator I also evaluated the thrust delivered by some modern propellers, having the same diameter as the ones the Wright Brothers said they had used in 1903, and they deliver enough thrust at 28 mph and 340 RPM.
There should be other more detailed articles, which I can not find, about the same set of tests because they planned the experiments for years and it is expected a lot of data was obtained.
I would be really interested to know either the exact geometry of the propellers to calculate their thrust with a simulator or find somewhere the diagram $Thrust
= Thrust_{340\:RPM}(v_{air})$ corresponding to the tested replica. If you can find such a document please tell me about it.
Anyway at 6 hp / propeller (12 hp in total) I calculated that only propellers with an efficiency well below 50% (miserable even for the standard of 1903) could have failed to produce enough thrust for the Flyer I replica at 28 mph.
Update 1
A Mathcad demonstration which shows that the efficiency of the propellers used by the Flyer I replica should be below 31.2% to fail the test T = D.

Update 2
Another wind tunnel set of tests was done in 2003 on a different Flyer I replica (see: Flying Qualities of the Wright 1903 Flyer:
From Simulation to Flight Test). Also the lift drag diagram for the case "props off" was about the same as the one obtained in 1999, the thrust generated by the propellers at 340 RPM and Q = 2 psf which translates in a 28 mph plane speed, the same as in the 1999 tests, was 80 lbf (36.28 kgf) more than double than the Drag = 17.44 kgf calculated for the 1999 replica. Unless quite different propellers had been used in 1999 and 2003 the 1999 propellers should not have failed the test T = D at 28 mph. It appears that various investigation teams have used propellers with different geometries.

Flyer I 2003 replica - Drag Coefficient, On Rail - Propellers off

Flyer I 2003 replica - Static and dynamic thrust of the propellers two dynamic pressures, $Q = 0.5\rho V^2$. 1) $Q = 1\:lbf/ft^2$ ($V=19.98\:mph$) 2) $Q = 2\:lbf/ft^2$ ($V=28.26\:mph$). $\rho=1.2\:kg/m^3$
Source: "Flying Qualities of the Wright 1903 Flyer:
From Simulation to Flight Test" (AIAA-2004-0105)

I do not understand. what is the question here?

The question is to find a more detailed article than "Full-Scale 1903 Wright Flyer Wind Tunnel Test Results" that will explain why the Flyer I replica failed the test T = D because as I already mentioned it could not have failed such a test unless the efficiency of the propellers had been well below 50% !! while the Wright Brothers maintained the efficiency was 66%.

they maintained a 66% efficiency where? in this other question is reported an extract from a letter where they write 23% http://aviation.stackexchange.com/q/21821/1467
also, I am not sure a "find this document for me" is a good question for this site.

Maybe you are more experienced than me in finding such documents. This is the reason I posted the topic on this site. 2) You gave me a link referring to something else. Somebody there mentioned 23% as being the efficiency of the gasoline engine used by the Wright Brothers not the efficiency of the propellers. 23% for a 1903 motor is quite good. That 66% is mentioned in a lot of places, example: http://wrightstories.com/propeller-design-demonstrates-the-genius-of-the-wright-brothers/

To clarify: Are you suspecting Nasa tests to have taken place on a replica which didn't conform to the original?

I added an update with more clarifications. 2) For the moment, I am not suspecting anything as long as I do not have the airfoil, the cord and the blade angle corresponding to a few cross sections spreading from $r=0$ to $r=R_{max}$. Using a free Calculator (see: http://www.drivecalc.de/PropCalc/) I get enough thrust at 28 mph with a constructed propeller that looks roughly like the original. The main issue for me is the pitch along the blade. I do not know it.

There's something we do not know. Replica Wright Bros flyers fail to climb out of ground effect, but the original did.

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Can a pendulum stabilizer for airplanes really work?
Can a pendulum really stabilize a plane for both roll and pitch?
I have read about various experiments (the one in the picture being just a simple case) but I still do not find an example of plane that was stabilised beyond any doubt with the help of a pendulum.
As a note:
If the suspension point of a pendulum is at rest or moves at constant speed then after some oscilations the bob will end up by hanging vertically. However this is not the case when the suspension point has an accelerated movement.

Source: Popular Science, December 1939
UPDATE
I have found a description (see the attached image) of a Wright stabilizer that was used successfully!! before February 1914. The roll was controlled by a pendulum (letter A in the picture) that swung along the main wings. At one point the article states that the device gave good results being tested daily for long intervals of time.
Another more recent text says that demonstrations were made before the Aero Club judges on December 31, 1913.

In the fall of 1913, Orville installed the stabilizer on a special
  Wright Model E airplane that utilized a single pusher propeller. He
  kept the details of the stabilizer secret even from the Wright
  Company. He purposely waited until the last day of the year to fly for
  the prize.
He invited the Aero Club’s judges to Huffman Prairie to see a
  demonstration of his new device on a cold snowy day, December 31st.
He turned up his coat collar, put on a pair of goggles and took off.
  He made a total of 17 flights.
His most spectacular flight consisted of 7 full circles of the field
  with both hands held in the air. The automatic stabilizer kept the
  same angle of bank and almost the same altitude. He wowed the judges
  and was awarded the prize on February 5, 1914.

Source: Wrights Develops Automatic Stabilizer
How is it possible?

Source: The Wright Automatic Stabilizer, L'Aerophile, February 1, 1914

A pendulum by itself seems strange. But an instrument based on an attitude indicator (or a conjunction of a gyroscope and a pendulum) may do the work.

As a child I saw a balsa and tissue model like this.  After trimming for level flight, the pilot tried a conventional hand launch.  The acceleration of course caused (the inertia of) the pendulum to activate full down elevator and the craft nosed dived to destruction.

Interesting find! Thank you for sharing your research! The pendulum works for pitch attitude control in unaccelerated flight only, and the fact that this is not being used widely after having been around for more than a century should give an indication of its limited usefulness.

As you noted, a pendulum can only "measure" pitch and roll angles when the pivot point is not accelerating.  Except, of course, if the length of the pendulum is the same as the Earth's radius — in which case we have a Schuler Pendulum.
Ed Carsten's pendulum stabilizer attempts to use the same principle as an autopilot measuring roll and pitch angles, and then generating elevator and aileron commands to level the aircraft. But this cannot work when using a pendulum of practical length since the roll and pitch angle "measurements" are in fact combinations of attitude angles and horizontal accelerations.

I suppose it is similar to trying to fly a plane by using a pendulum hanging from the cockpit ceiling as your vertical reference.

@ Christo, This is what Orville Wright did on December 31, 1913..

If the pivot (suspension point) of a pendulum is inside a plane, for example in the ceiling of the cockpit, it does not matter were, it will have a random acceleration and in consequence it will move like in the animation you can see below. The demonstration can be found here. The pendulum never aligns to the vertical and so it is useless as a reference able to indicate which way is down.

Pendulum with a moving pivot. The small circle is the pivot, the big circle the bob. If the pivot has a chaotic movement the pendulum also follows a random trajectory.
Source: Pendulum with a moving pivot by Carl Madigan

The Wrights' automatic stabilizer used a pendulum for control of the roll axis only. Control of the pitch axis was based on a vane to sense angle-of-attack. As far as the roll axis is concerned, a pendulum gives the same information as does a slip-skid ball (though a pendulum is less damped and more prone to unwanted oscillations).  In general, when an aircraft is banked, it tends to sideslip.  In more detail, when an aircraft is banked, the flight path curves and as a result, the outboard wingtip tends to experience more drag than the inboard wingtip, so the nose tends to point toward the outside of the turn, unless the rudder is deflected as needed to prevent this. The resulting sideways airflow over the aircraft creates an aerodynamic sideload that tends to displace a slip-skid ball toward the low wingtip, and likewise a pendulum. So the principle of using a pendulum for roll control is theoretically sound, based on the relationship between bank angle and sideslip. This relationship between bank angle and sideslip is also at the core of why dihedral tends to roll an aircraft towards wings-level-- with no sideslip, dihedral would generate no roll torque regardless of the bank angle.
It would seem that the Wrights could have used a vane to sense sideslip directly instead of a pendulum.
It also would seem that that Wrights' stabilizer system might work best on aircraft with counter-rotating propellers, where confounding factors such as p-factor are eliminated.  However the stabilizer system was in fact initially demonstrated to the public on Wright Model E with a single propeller-- see http://wrightstories.com/wrights-develop-automatic-stabilizer/
The relationship between bank angle and sideslip is subtle enough that neither a pendulum nor a slip-skid ball--or for that matter, a yaw string or other sideslip sensor-- would make a practical cloud-flying aid, even speaking strictly in regard to the roll axis alone, except perhaps in the most absolutely ideal conditions. In very smooth air it might be possible to get some meaningful information from a slip-skid ball or pendulum or yaw string while flying blind if the pilot uses certain unconventional control inputs-- perhaps best results might be obtained by limiting the aileron deflections to extremely small angles, while leaving the rudder centered. Other similar variations on the same theme come to mind as well. This is not something that can be recommended as a practical technique in the real world.
Note that any system of blind flying that uses the slip-skid ball in combination with one or more gyro instruments operates on completely different principles than are described above, even if-- as in the obsolete and confusing "1-2-3" system of blind flying-- the pilot is instructed to use the ball as the primary guide to his aileron inputs.  See Langewiesche's "Stick and Rudder" for commentary on this antique method of blind flying.

A pendulum could be used to sense (average) pitch. Dampening would be required to avoid oscillations. But then, what would be the point? I understand that the objective was a safe angle of attack. A simple preset of the elevator would perfectly achieve that.
The pendulum cannot work for roll. A banked plane will follow a curve. The sum of forces will remain perpendicular to the wings. Only changes in the roll rate will affect the pendulum, which is not useful for a simple mechanical control setup. (Imagine sitting in an airliner, far away from a window. You can feel the start and the end of the roll, but not the roll itself)

How would the pendulum work for pitch if the aircraft decelerates?

@Koyovis A pendulum cannot tell the difference between nose down pitch and deceleration.

Indeed dynamic behaviour of the pendulum is such that it is useless for the purpose of stabilising the aircraft.

No, a pendulum wouldn't work, mostly because it would be too sensitive to spurious  accelerations. Imagine a system like that in turbulence, the pendulum would start swinging around and would create unpredictable control inputs which would need to be damped out by the pilots.
Mechanically, you would need to have a pendulum arm as long as possible, this would take up a lot of space you would rather use for other things like passengers. It would also probably be fairly heavy, reducing useful load.
Lastly, there are much better systems out there. Why would you use a heavy, bulky mechanical pendulum for one axis when these days you can buy an all singing and dancing 3 axis autopilot on a single chip with redundant laser ring gyros, accelerometers and GPS?

Can a heavy pendulum really work, at least in theory?

Those autopilots on a chip don't use laser ring gyros, but MEMS gyros which measure rotation on the basis of the Coriolis force.

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text:
Can the plane Celera 500L be eight times more fuel efficient than a jet of similar size and capacity?
Celera 500L (see Otto Aviation site for more details) is a plane with a single propeller powered by a piston diesel engine. The constructor claims it is 8 times more fuel efficient than a jet of about the same size and capacity. How is this possible?
Another question would be: Is this flying apparatus at least twice as efficient as the best piston engine plane of comparable characteristics?

Otto Aviation - Celera 500L - V12 turbo-diesel laminar flow plane
General characteristics

Capacity: six passengers
Powerplant: 1 × RED A03 diesel piston engine, 550 hp (410 kW) approximate at takeoff

Performance

Cruise speed: 400 kn (460 mph, 740 km/h) estimated minimum
Range: 4,500 nmi (5,200 mi, 8,300 km)

Claimed comparison between Celera 500L and a Jet
"[Celera 500L] is a six-person private craft that promises to fly at jet speeds, but with eight times lower fuel consumption, and a range that's twice that of a comparably sized craft."
"The company, founded in 2008 and an offshoot of Bill Otto's Otto Laboratories, says that the Celera 500L runs at 18 to 25 miles-per-gallon fuel economy (compared to the 2-3 miles-per-gallon of a comparable jet aircraft).
"Then there are the modest $328 hourly operating costs, which are about six times lower, and the generous 4,500-nautical-mile range. Maximum cruise speed is projected to reach more than 460 miles per hour."
"The reason its aircraft can do all this, says Otto Aviation, is down to laminar flow.
"Laminar flow is the minimum drag solution for aircraft surfaces, explains its website, and features smooth layers of airflow with little to no mixing of adjacent layers."
"With its aerodynamic airframe meaning it requires a lot less horsepower to achieve takeoff and cruise speeds, the Celera 500L is powered by the RED A03 engine. It has a Liquid cooled V12, twin 6-cylinder bank and, says Otto Aviation, offers best-in-class efficiency. It's certified to operate on Jet A1 and biodiesel."
Source
UPDATE:
According to a patent granted to William Otto (US9669939B2, "Aircraft supplemental thrust device and method of operating the same") Celera 500L has a service ceiling of 19.8 km!

"Propulsion of the aircraft may be provided by a fixed-pitch eight
blade composite blade propeller mounted at the rear of the fuselage on
the centerline axis. The propeller airfoil sections and section
incidence angles are configured to provide maximum efficiency at
cruise at 50,000 ft. altitude and above. Propeller diameter is also
optimized for the high altitude cruise environment and as a result
essentially eliminates supersonic blade velocities during low altitude
operation. The optimum propeller diameter is slightly smaller than
maximum fuselage diameter which coincidentally reduces the probability
of bird strike and other foreign object damage.
...
The aircraft cabin may be approximately 74 inches high and include an
approximately 78 inch width having a minimum 50 inch seat pitch. The
aircraft has a service ceiling of approximately 65,000 feet, and a
normal cruise speed of between approximately 460 to approximately 510
mph, with a specific fuel consumption of approximately 30 to
approximately 42 mpg depending on cruise speed and altitude. Landing
stall speed is approximately 70 mph, takeoff and landing speeds are
approximately 90 mph, and runway requirements are approximately 3000
ft."

I have seen on Otto Aviation's site this text:

"The Celera 500L has a glide ratio of 22:1 (typical GA aircraft of
similar size have a glide ratio of < 9:1). At an altitude of 30,000ft
The Celera 500L can glide up to 125 miles with no engine power. This
is roughly 3x better than the typical aircraft."

and I believed that the projected service ceiling of the plane is 9.144 km not somewhere between 15.2 km and 19.8 km. At such high altitudes the air density is between 0.2 and 0.1 kg/m^3 and in consequence Celera can, in theory, reach 740 km/h with 550 hp. The Grob Strato 2C, a German experimental high altitude research aircraft, powered by two turbocharged piston engines, reached a record altitude of 18552 m on 4 August 1995, with two 6 m in diameter propellers, using a total of 800 hp.
Question: Is this altitude of 65000 or at least 50000 feet achievable by Celera 500L?

Only if it flies very slowly. If compared on equal footing the answer is a clear no. Current airplanes are also quite refined designs.

I would suggest editing the question down to only that which is relevant to the actual performance claims you are questioning.  A link to the article would be nice for those who are interested in learning more, but there is a lot of extra marketing fluff you chose to copy/paste that has nothing to do with your basic question.

@MichaelHall, I can eliminate 2-3 paragraphs but the rest of the text is important information that represents in fact technical and economical characteristics of Celera 500L.

It looks like Celera 500L is an investment scam.

@Robert Werner, I counted around 11 paragraphs that I consider nothing more than company PR.  Don’t get me wrong, it’s interesting to aviation buffs, but not pertinent to the question.

That looks a lot like a P180

Looks like they are trying to extend the laminar flow region over the fuselage by using the Goldschmied propulsion method. There are papers which claim some additional thrust from this technique.

That thing looks extremely claustrophobic for the passengers who get nothing more than a glimpse outside through the cockpit windows.

@PeterKämpf I added an update to the question because I found a patent of William Otto where the projected service ceiling of Celera 500L is taken as 65000 feet or at least above 50000 ft. Is such a range of altitudes achievable by Celera?

@RobertWerner This depends on the turbocharger of the engine. I would expect that it has a critical altitude of not more than 8000 m, most likely less. In order to work at full power in 20,000 m (65 kft), I would expect it to have two turbocharging stages (with humongous intercoolers which clearly the 500 L doesn't have) plus one supercharger.

@Peter Kampf the Grob Strato 2C also carried a 2700 horsepower compresser.  So we begin to see the issues here. The 2 meter props are on the right track.  Although the Grob was heavier, the additional power requirement is undeniable.

@RobertDiGiovanni Grob Strato 2C uses: 2 × Teledyne Continental TSIOL-550 , 300 kW (400 hp) each and 1 × Pratt & Whitney Canada PW127 gas generator. However, there is no word about the power needed by this gas generator to feed the piston motors with enough air to make them generate 800 HP. The power can not be 2700 HP. (see: https://en.wikipedia.org/wiki/Grob_Strato_2C ).

I am seeing that the engine builder intends to make a motor that can work up to 50000 ft, also the max power can be reached only below 25000 ft. [RED A03 200 Series: Increased Take-Off Performance - Derivative of A03-003 - Designed for Very High Altitude Application - Provides Full Take-Off power up to 25.000ft - Operational Altitude up to 50.000ft - Twin Stage Turbocharger Induction System (Source: https://red-aircraft.com/product/)]. Definitely the engine Celera is using now is not capable of working at 15 km above the sea level.

@Robert Werner the P-51 Mustang has a glide ratio of 15:1 at 175 mph.  The straight wing pusher design, and its champion, the 230 mph B36 Peacemaker, disappeared rapidly in the 1950s when designers found they could push faster with jets.  Swept wings, fan jets, and supercritical wings sealed its fate.  The Mustang burns 100 gallons per hour at 300 mph.  The 747 burns 5 gallons per mile, but at 400 tons = 80 ton miles per gallon! (size helps too) The jets cruise at twice the speed.  I look forward to flight data as they progress, with possibly an application as a shuttle craft between airports.

tl;dr: L/D of the airframe and specific fuel consumption of the engines are credible but that doesn't add up to the claims made regarding fuel consumption and speed relative to jets. Laminar flow seems to be used as the snake oil to sell this to gullible customers.
The fuselage shape is inspired by a laminar flow airfoil, like the NACA 64-021 shown below, courtesy Airfoiltools:

Now let's have a look how much laminar flow will be possible. Flight speed is claimed to be 205 m/s in maybe 25,000 ft = 7620 m. The Reynolds number per meter in these conditions is 7,314,000. Flat plate flow shows laminar to turbulent transition at about 400,000. With a stabilizing pressure gradient this can be shifted to maybe 4,000,000. So the first 55 cm of fuselage length will exhibit laminar flow but then turbulent transition is unavoidable. I cannot find dimensions for the Celera 500 L, but I think it is fair to say that laminar flow on the fuselage is insignificant. This corroborates well with other aircraft for which a lot of laminar flow had been claimed.
Still, the very clean shape gives it a clear advantage. Add to that high aspect ratio wings with laminar flow, and Otto Aviation's claimed L/D of 22:1 is entirely credible.
Now for the engine: Diesels achieve 220 g/kWh since decades. RED gives 210 g/kWh for the A03, which is again entirely credible. The comparison with turbine engines is complicated by the fact that they define specific fuel consumption per unit of thrust, so we need to look at the engine-propeller combination at 205 m/s. The propeller on the Celera 500 L has five blades and a rather small diameter for take-off rotation and to keep tip speeds subsonic, so its efficiency will maybe be 82%. If we run the A03 at 92% which is its stated maximum continuous power of 338 kW, thrust will be 1352 N and fuel efficiency will be 52.5 kg/kNh which is the same number as what the GEnx-1B64 achieves in cruise. Let's be realistic and double that number for a regular small jet engine (Published numbers are for static conditions and cannot be directly compared. Doubling them for cruise conditions is a good approximation. These numbers are representative of older jet engines and grow by a factor of maybe 1.5 for cruise).
Now we need L/D figures for small business jets. This source doesn't cover jets but shows that small land airplanes with retractable gear still achieve values between 13:1 and 18:1, so even a combination of a 1950's jet engine and an average airframe will only have four to five times the fuel consumption compared to the Celera 500 L. Four, not eight!
For an L/D of <9:1 the comparison airplane needs a fixed gear or floats. With standard design and built any competitor should at least get 13:1 or better.
Sanity check: If the 22:1 L/D can be sustained up to cruise speed, the whole aircraft cannot weigh more than 3033 kg, which is less than half of a King Air 350 which Otto Aviation uses for cabin size comparisons. The King Air, having a smaller cabin and lighter PT6A engines, still needs an MTOW of 6800 kg. Judge for yourself …
Otto Aviation has done everything right, engine choice, propeller position, wing and fuselage layout all support the lowest possible fuel consumption. However, the existing aircraft aren't so poorly made in comparison, either. Also, the installed power looks impossibly low for the claimed cruise speed. In order to fly most efficiently, the Celera 500 L needs to cruise at a much lower speed.
Now for the maximum altitude: No, the Celera 500 L will not reach 50,000 ft. Not even close.
I know a bit about the Strato 2C. It was too heavy for its intended altitude of 65,000 ft. Wing loading needs to be low enough and the engine has to have enough turbo- and/or superchargers to lift its critical altitude high enough. In case of the Strato 2C, it used two turbocharger stages with inter cooling and one supercharger. The intercoolers determined the size of its engine gondolas, which were huge. Given the small air intakes of the Celera 500 L, it looks like it will only reach 25,000 ft or maybe 30,000 until cooling airflow becomes insufficient to keep the intercoolers supplied. Unless Otto straps a rocket booster to its tail, it will never go to even 50,000 ft.

You mentioned "two A03", doesn't this contraption have only one?

@Jpe61 As far as I understand, it has two acting on one gearbox and propeller. Similar to the LearFan 40 years ago.

Judging by the image material available the RED A03 has a single cranckshaft, so even if each 6 cylinder bank is capable of independent operation (I guess this is possible considering vibrations even in long term), and it is advertised to have other redundant features, It definitely is a single engine. All the stuff I found about this Celera (which I consider a funding hoax) states a single RED A03 powerplant.

@Jpe61 With a single A03 the claimed speeds are completely impossible. And yes, there is only one so I have to rewrite this answer. Your funding hoax opinion looks more and more true.

I suspect fraud.  The performance figures these guys are claiming just don’t seem possible with this kind of an airframe.

Investing in aviation offers excellent chances for spectacular losses. I've had my fair share of them

The Long EZ is said to get 30-40 miles per gallon.  But slower and much lighter.  But they could hook up with Rutan and maybe get some windows on that thing.

I'd like to know how it performs in rain. Based on personal history with a friend's Quickie2, laminar flow wings and rain are not a good combination. If this is to be a business aircraft, it needs to be all weather capable.

The Long EZ doesn't get such milage in real life, especially at high speeds. There are one-off examples (Gary Hertzler, Klaus Savier and others) who have tweaked the smaller and lighter Vari-Eze to the max to get as much as 42 MPG but not at 200 MPH, let alone 460. Their efforts and results in the ten CAFE400 air efficiency races are still remarkable.

You doubt the efficiency of Williams engines, Peter? Those figures are actually accurate.

@JWalters: Wikipedia doesn't even give a speed for which the SFC is valid. How should I trust that?

@ymb1 This is the length after which a Reynolds number of 4 million has been reached and where I assume transition is likely.

The following claim was made:
"Otto Aviation explains that the Celera 500L enjoys a 59% reduction in drag when compared to similar-sized airframes." (Source)
I will calculate the Power needed by Celera 500L to fly at 740 km/h assuming its drag is 1-59% that of a Cessna 340, a comparable plane.

Cessna-340 - Crew: one pilot; Capacity: five passengers: Powerplant: 2 × Continental piston engines, 310 hp (230 kW) each; Maximum speed: 244 kn (281 mph, 452 km/h); Range: 1,406 nmi (1,618 mi, 2,604 km)
For two planes 1 and 2, if 1 has 59% less drag than 2 then the following equations can be written (1=Celera 500L and 2=Cessna 340):

Regarding the fuel consumption in miles per gallon (see the second part of the calculations above), as the diesel engine of Celera 500L has a SFC = 210 gr/kWh and diesel oil has a density of 0.85 kg/l, it results that, at the speed of 740 km/h (engine power = 1115 HP), Celera 500L has a MPG = 8.47 mi/gal
If we are to believe this statement:
"A 17,000 pound Lear Jet 35, capable of carrying seven people at 485 mph, gets about 4 mpg",
then Celera 500L is about 8.47/4 = 2.11 times more fuel efficient than a Lear Jet 35, not 8 times, and it needs an engine of at least 1115 HP to fly at 740 km/h.

It would have to make around 275 mph indicated to get a TAS in the mid 400 mph range at 8000m.  One might wonder as this was the very best a P-51 Mustanger could do with over 3x the rated horsepower.

The lowest drag of an airframe is possible when zero-lift drag is half of total drag. So I propose to double your drag number to account for induced drag and the maximum possible MPG figure in flight is only half of what you use.

@PeterKämpf  You are right. I have to rewrite my answer to be more explicit. The 487 HP I calculated is the min theoretical power necessary to make a symmetrical revolution body, with a C_D=0.052 and a max cross section = pi*1m^2, travel at 740 km/h through an air with the density=0.52 kg/m3 (8000 m). My C_D is not the Drag Coefficient used in aviation but an ordinary drag coefficient and the body has such a weight as to fall nose down at 740 km/h. There is no lift, only drag. I wanted to show that Celera can not travel through the air at 740 km/h with less than 487 HP no matter what you do.

@PeterKämpf So, basically, Otto Av says Celera has about 59% reduction in drag as compared with a quite similar plane, LearAvia Lear Fan 2100 (see: https://en.wikipedia.org/wiki/LearAvia_Lear_Fan ), that never entered production but the builders claimed for it a max speed of 720 km/h with a total power of 960 kW. If I apply the formula in the answer above I get Power_Celera = (1-59%) x 960 kW x (740 km/h / 720 km/h)^3 = 573HP. There are two questions here: (1) Are the claimed characteristics of Lear Fan credible? (2) What Otto Aviation did so that it obtained Drag Celera = 41% x Drag Lear Fan?

@RobertWerner: This is hard to answer in a comment. The LearFan was a honest attempt to build the most frugal business plane possible but FAA hobbled the carbon fiber design and the gear box kept overheating. But with 2 PT-6 it did reach the claimed speed. Otto will not. So (1) is yes and (2) is most likely not true but pure Marketing.

I like your answer, but please next time save the equation graphic as a .png rather than a .jpg.

@rclocher3 Now the Mathcad file with formulas and calculations are in .png format. I have changed it.

Thanks @Simplex11!  That's easier on the eyes.  (I'll delete my comments in a little while since they're not relevant any longer.)

To get in the ballpark, one may look at the Predator, Reaper, Avenger drone evolution.  Piston prop, turboprop, fan jet progression for more speed using very efficient high aspect wings.
A useful efficiency comparison metric amoung transportation vehicles is ton mile/gallon.  It is important to note 460 mph aircraft (275 mph indicated) weighing 3 tons that get "18 - 25 miles per gallon" simply do not exist.  Not even close.  This is the most dubious claim.  18-25 ton miles per gallon is more like it.
So how do "gas guzzling jets" do well in this competition?  The ability to continue generating thrust at higher altitudes.  Piston engines struggle to get enough air and do not generate enough horsepower to effectively turn both the prop fast enough to generate adequate thrust and  run the compresser at higher altitudes Adding to their woes is the reduced efficiency of the prop as forward speed increases.  Additionally, in thin air, there is little doubt that the tiny Celera prop must go supersonic.  A 550 horsepower piston engine doing all that? Not likely.  The aircraft in its current configuration may reach 200-250 mph TAS.
Jets shine at higher altitudes in the "tons" and "miles" categories even though they are worse in "gallons", taking advantage of their ability to provide adequate thrust in thinner air.
Much to my surprise, the 450 mph Avenger drone came in at 50 ton mile/gallon, while the slower turbo prop Reaper manages around a fairly typical 25.  An improvement on 25 at Reaper airspeeds is what we may expect from the Celera 500L.
Generating enough thrust high will be the challenge with this design, giving them credit for reducing fuselage drag.
Amazingly, compression of an adequate mass of air to generate sufficient power to fly high ... is what jets do best.  I would be looking at those drones.
Some more efficiency data to consider:  XB 35 flying wing (no fuselage/ contra rotating pusher props) cruise speed: 240 mph ton miles per gallon: 75.  B36 Peacemaker (conventional design/ pusher props) cruise speed 230 mph ton miles per gallon: 55. B52 Stratofortress (8 turbojet engines) cruise speed 525 mph ton miles per gallon 40.  Boeing 777 airliner (2 high bypass fan jet engines) cruise speed 560 mph ton miles per gallon: 58
U2 spyplane (F104 with very large wings) cruise speed 430 mph ton miles per gallon 46 (a little smaller at 40,000 lbs).  Cessna 340 cruise speed 210 mph ton miles per gallon 21 (6000 lbs).
Conclusion:  this design may offer improvements in fuel efficiency over similar sized prop driven piston engined aircraft.  The major concern would be landing speed and the ability to climb to higher altitudes.  However, it may find a commercial niche (manned or unmanned) as a shuttle between large airports with longer runways.

The rear propeller and the part of the vertical tail placed between the axle of the propeller and the ground (I guess to protect the tips of the blades from hitting the ground during landing and take off) do not seem a good idea for a manned plane. If the vertical tail hits the ground it can be easily damaged and the propeller itself can have the same fate.

The vertical tail issue can be worked around (with longer mains).  At 8000 m the very best props make around 40 ton miles per gallon.  At 4 tons, they may get around 10 mpg at around 300 mph.  They may have scaling issues with their visions, but this is a great experimental project and I wish them luck.

A Cessna-340 (1 pilot and 5 passengers), powered by piston engines with 620 HP total power, can not go faster than 452 km/h. In consequence, to reach 740 km/h (the speed of Celera 500), a Cessna-340 would need 620HP*(740/452)^3=2720HP. Celera 500L claims maximum 550HP for 740 km/h!! This is not credible.
[Cessna-340 - Crew: one pilot; Capacity: five passengers: Powerplant: 2 × Continental piston engines, 310 hp (230 kW) each; Maximum speed: 244 kn (281 mph, 452 km/h); Range: 1,406 nmi (1,618 mi, 2,604 km). (see:https://en.wikipedia.org/wiki/Cessna_340)]

A 737-400 gets about 32 ton-miles per gallon in best cruise. 436kts 150k pounds(filled up). based on an actual operations manual. These are US short tons(2000 pound) and proper nautical miles. (statute miles haven't been used in legal aircraft specifications, like operators manuals, since the FAA forced better standardization around 1976)

Not really significant at cruise, but it would be correct to use CAS not IAS for aero calculations. IAS could hypothetically be any arbitrary number based on equipment calibration, plane configuration, and static air source; but CAS is always just physics.

@Max Power yes, thats why they wanted the more efficient engines on the 737.  Size (mass to surface area parameter) favors larger aircraft.  32 ton mile per gallon is still very good, especially for a jet.  The Cessna 172, by comparison, is around 11.

You can just edit your answer with a footnote to clarify the units used. The 737-400 is second gen not an NG(3rd), the NG had a full airfoil change. I get about 14-16 ton-mile per gallon in cruise out of the c172s I fly.(nautical miles)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Does oxygen become less available during night-time?
The following excerpt is from the FAA's Pilot's Handbook of Aeronautical Knowledge (page 7-37):

Some people flying above 10,000 feet during the day may experience
  disorientation due to the lack of adequate oxygen. At night,
  especially when fatigued, these effects may occur as low as 5,000
  feet. Therefore, for optimum protection, pilots are encouraged to use
  supplemental oxygen above 10,000 feet cabin altitude during the day
  and above 5,000 feet at night.

The book, in vague terms, suggests that the amount of oxygen (i.e. oxygen molecules) available at a certain altitude at night is less than in daytime.
The regulations do not differentiate between day and night in regards to supplemental oxygen.
What would be the scientific reasoning behind this?

I have to wonder about this, since in most of the places I've lived during my adult life, I'd have to dig a hole - often a pretty deep one - to get to 5,000 ft.

@jamesqf: If you live above 5 kft AMSL, you're almost certainly acclimated to slightly higher pressure altitudes than sea-level dwellers, and you personally can probably get away with using a somewhat higher altitude cutoff than what the FAA suggests.  (Disclaimer: IANAD.)

Night vision is one of the first things affected by reduced oxygen levels, and that effect is masked in daylight, but can become consequential at night.  But no, the O2 molecules are the same night or day.
BTW, airline pilots in our 8000' cabins do not have different guidance for putting on the masks -- it's 10k, day & night both. Air Force was, IIRC, the same 10k, with no difference in the rules day vs night.

Primarily because of this physiological effect, I use supplemental oxygen at night above 5,000', regardless of the regulations.

And likely it's because we don't use the same eye function at night, the night vision is primarily based on rod cells. Rods are already working at an incredible sensitivity, lack of oxygen quickly destroys this sensitivity.

The question was edited by @pondlife and unintentionally changed I believe. I never meant to "ask" if FARs have different regulations for day and night. I was stating the fact that no such regulations exist. I've re-edited the question now to make this clear.

Humans are diurnal. At night, we get sleepy. The effect is worse when we're extremely tired.
As a result, we are more susceptible to things that affect cognitive processing during those times. Oxygen inadequacy or deprivation (meaning a reduction below the level we are physiologically accustomed to) is one of those things.
It's nothing to do with the actual amount of oxygen at a given altitude changing at night, just our ability to ignore the effect of the existing changes compared to ground level when it is night time.

This answer would benefit from supporting scientific references.

The amount of oxygen in the air is the same day or night. The big difference - which isn't really explained in the part of the PHAK you quoted - is that at night your eyes work differently.
Your eyes have two different types of photoreceptors: rods and cones. Cones work best with high light levels and rods work best at low light levels. So during the day, when there's plenty of light, you're seeing primarily with the cones. At night, it's the opposite: your vision depends mostly on the rods.
But, rods are very sensitive to oxygen levels. As the level decreases, they quickly stop working effectively. During the day this doesn't matter because you're not using them anyway, but at night it does matter and you can lose night vision at relatively low altitudes. This is why the PHAK recommends supplemental oxygen at night: it's to make sure that the rods continue to work effectively.
The PHAK has a lot of information about "vision in flight", starting on page 17-19. Page 17-24 explains the need for oxygen at night:

Unaided night vision depends on optimum function and sensitivity of
  the rods of the retina. Lack of oxygen to the rods (hypoxia)
  significantly reduces their sensitivity. Sharp clear vision (with the
  best being equal to 20–20 vision) requires significant oxygen
  especially at night. Without supplemental oxygen, an individual’s
  night vision declines measurably at pressure altitudes above 4,000
  feet. As altitude increases, the available oxygen decreases, degrading
  night vision. Compounding the problem is fatigue, which minimizes
  physiological well being. Adding fatigue to high altitude exposure is
  a recipe for disaster. In fact, if flying at night at an altitude of
  12,000 feet, the pilot may actually see elements of his or her normal
  vision missing or not in focus

Somehow I suspect the PHAK was written by/for sedentary people living near sea level, rather than the (rather large) fraction of the population who've lived most of their lives above 4,000 ft.  Not to mention those of us who enjoy spending days hiking up mountains 12K ft or more :-)


text:
Why do stealth aircraft generally lose in maneuverability?
Whilst stealthy built aircraft have (or claim to have) many anti-detection systems, they seem to have to compromise on less maneuverability features, and even have to keep their missiles inside, not on the outside.
Why is it so these days?
Based on the above what could be the tendency for the future military aircraft (the jet fighters in particular)?

this is a very similar question I have just seen now:

Some stealth aircraft are bombers, do you want to discuss only about jet fighters right?

I was really only thinking about the the jet fighters

Why do you assume they lose maneuverability? The F-22 is certainly not less (if anything, it is more) maneuverable than an F-16. The fact that the armament is kept inside contributes to maneuverability, not harms it.

I am not sure between F-22 and F-16, I was comparing for example the F-35 with the Sukhoi-37. when stealthy is the subject the one that comes to my mind is the F-35.

Then stealth has nothing to do with it. You are comparing apples to oranges. The SU-37 is much more maneuverable than many other fighters, whether or not they happen to be stealth is irrelevant

I thought that to make them stealth they had to compromise on the manoeuvrability

Radar stealth is achieved by carefully designing the aircraft to reflect radar waves in specific directions. It helps to keep the aircraft shape as simple as possible.
External weapons make this more complex: each missile adds a large number of surfaces that can reflect radar. The missile can also interact with the airplane fuselage to reflect radar waves several times. And once the missile has been launched it leaves an empty pylon behind with yet more surfaces capable of reflection.
It's much easier to maintain stealth if the missiles are stored inside the aircraft.
Some aircraft (e.g. the F-117) also try to achieve IR stealth by hiding the exhaust, mixing the exhaust gases with air etc. This is difficult to combine with vectored thrust.
Still, an aircraft like the F-22 has pretty good maneuverability, thanks to good aerodynamics and huge amounts of engine power.
The Su-37 uses vectored thrust and canards (again, extra reflection surfaces) to achieve high maneuverability.
So, there's a tradeoff going on. Do you want to blow your enemy out of the sky before he's seen you (which requires stealth)? Or do you focus on close combat (where maneuverability is most important)?

Short answer:  There is no reason that a low observable aircraft should suffer in maneuverability, if properly designed, and the F-22 and YF-23 are proof of concept.
LO does favor specific shapes which both minimize the amount of radar energy reflected back to a transceiver as well as provide good aerodynamic qualities.  Weapons carriage must be done internally in order to prevent compromising this shape and increasing the aircraft's RCS.  LO aircraft also possess an additional advantage over existing airframes with external weapons carriage in that internal stores carriage creates far less parasite drag than external carriage does, providing more excess thrust for maneuvering and accelerating at any given airspeed.
Again, if an aircraft is properly designed for a particular role and maintains strict design discipline and focus on the goals for the program, there is no reason a stealth airplane cannot possess good fighter qualities with low observable features included.  F-117 and B-2 are not terribly good comparisons on this issue as both were designed as bombers, though a B-2 does demonstrate stellar aerodynamics blended with LO features, having an Mmo of Mach 0.95, which is unheard of in the civilian world.  F-35 was a bloated compromise between three variants with cripples it in terms of traditional fighter metrics and is proof not that LO aircraft don't make good fighters but that multirole aircraft are not going to be stellar in any specific mission role.

The answer may lie in change of tactics.  Stealth aircraft are all about one of the oldest rules of gaining advantage:  don't be seen first.  This is combined with the modern:  a 9g plane can not evade a 25 g missile.
They want the other plane trying to evade your missile.  The compromises in aircraft performance are understandably lamentable, but being "not seen first" remains the key to survival.
However, technology and tactics are always changing, and it may be wise to remember that many our Air Force's greatest successes came from teamwork of different types of aircraft working together, rather than putting everything into one plane.
Once a slower stealth is located, it then becomes a plane v plane situation.


text:
Where do flaps and elevators get input from?
How are flaps and elevators controlled in both a fly-by-wire or manual airliners? Is it from the yoke or the joystick as in Airbus or military aircraft?
For how long and in what stage of the flight are flaps kept down? E.g. at takeoff, are flaps down until the aircraft reaches the required altitude or do aircraft rely on just elevators-down-nose up position to change altitude or raise the nose?

Let's take the two most prolific airliners as examples, the A320 and the B737.
A320
For aircraft pitch control, the A320 has a horizontal stabiliser which can be trimmed, and an elevator.

The elevator can deflect very rapidly, and is actuated via hydraulic actuators. They receive their inputs from the digital flight computers, which compute appropriate inputs depending on pilot side stick input and aircraft flight state.
The horizontal tail can be trimmed: moved relatively slowly via a screw jack driven by two hydraulic motors. The hydraulic motors receive their inputs from the flight computers, which provide commands for auto-trim of the stabiliser. Direct mechanical control can also be used: the trim wheels have a mechanical cable connection with the hydraulic motors, rotating the trim wheels directly commands the hydraulic motors to run. The trim wheels are situated on either side of the thrust levers.

Flaps get their input from the flap lever. The A320 simply has settings 0,1,2,3,Full, and the pilot selects the proper number in the appropriate flight phase. From A320_Briefing_For_Pilots:

B737
The B737 has the same arrangement as the A320: a moveable horizontal stabiliser, and an elevator.

Again, the elevator makes the rapid deflections, actuated by hydraulic actuators. They receive their inputs from mechanical cables, connected directly to the columns in the cockpit. A feel spring provides artificial entering forces, proportional to the airspeed (q-feel).
The stabiliser is actuated via a screw jack, which can be rotated via electrical motors (by flipping the trim switch on the control wheel) or directly via the trim wheels, which are connected to the stabiliser via mechanical cables. From this link:

Trim can be applied by electric trim switches, autopilot or a manual trim wheel. Electric and autopilot trim may be disengaged by cutout switches on the control stand in the event of a runaway or other malfunction.
Moving the control column in the opposite direction to electric trim will stop the trim, unless the STAB TRIM switch is set to OVERRIDE. This function could be used to control the pitch of the aircraft with trim say in the event of a jammed elevator.
The trim authority varies according to aircraft series and method of trim. The full range is only available with the manual trim wheel, but if at an extreme setting, electric trim can be used to return to the normal range. There are two electric trim switches on each control column, the right is for the direction and the left is an earth return for protection against spurious electrical signals.

Flaps again get their inputs from the flap lever, next to the thrust levers. There are slots for the lever to be situated at the appropriate flap setting, noted in degrees deflection.
Image source

In most airliners the flaps are controlled from the flaps lever. It is usually in the central pedestal ("throttles quadrant"), near the throttles.
(right most handle that says "flap")

Courtesy of www.b737.org.uk

Flaps usage is usually dictated by speed, not altitude (although some airports restrict flaps usage below a certain altitude due to noise abatement regulations, for which NADPs are published, for example). The lower the speed, the lower the flaps would be.
The change of the nose attitude is not why the flaps are used. It is merely a byproduct and it must be anticipated and corrected by the pilots (in most aircraft the flaps create a nose pitch down moment, but in some designs the nose may pitch up).

If I am not mistaken, flaps usually create a pitch down moment (consistent with center of lift moving aft)...?

@CptReynolds you're correct, I'll edit

In some planes flaps do both ;) In the C172 I'm training in, the first 10 degrees of flaps pitches the nose up while 20 and 30 degrees pitches down.


text:
What ATC entities does a commercial airliner communicate with?
As a private pilot, I am familiar with tower, ground, approach and departure control, but I am aware that there are other entities like clearance, pushback and center. If I were flying a commercial airliner, what is a complete, ordered list of entities (and a brief description of roles) I might talk with over a flight between two class B airports?

Even as a private pilot, you should know clearance delivery, and center, maybe even talk to them once in a while :)

Fair, @roe. But these things atrophy. :/

Commercial airlines communicate with the same ATC entities as private planes, although some routes and frequencies may be different based on speed and altitude. In addition, they communicate with their company.

@roe, in the US, clearance delivery is usually limited to IFR flights. Only at truly busy class C and B airports will VFR flights get squawk codes on departure, and even then that is often handled by Ground.

egid; true, and private pilots with an instrument rating, would be using said clearance, no? That said, there are airports which require you to contact clearance first (like Stuttgart in Germany, although they use the call-sign 'delivery' instead of 'clearance', which is kind of weird..), even to just tell them you want a northbound VFR departure or whatever. But yeah, I guess clearance is a bit less common.

If anyone could enlighten me to how often and for what airliners use the company frequencies though, I'd be much obliged, been asking that myself a couple of times. Is it for reporting time of departure and ETA and stuff?

That's true. I do feel like most private pilots in the US are not instrument rated, but I admit I overlooked that :)

Using clearance VFR is pretty rare in the US.  I've never done it.  Only when IFR.  Not all the frequencies are always in use either.  The towered fields I visit usually have the tower controller doing ground too and monitoring both frequencies, but just normally transmitting on one (on landing tower will say "stay with me for ground", on startup the initial ground call will be returned with instructions to switch to the tower frequency, presumably to minimize the controllers frequency hopping)

@BrianKnoblauch At sleepy little KISP they make VFR GA flights talk to clearance first - They may be trying for the "emptiest Class C airspace in the world" award, but by the administrator's bones they're using it for all it's worth! (I think it's because they have an inferiority complex what with JFK & LGA right to the west...) Most of the class D fields around here have clearance & ground combined though and sometimes it's just one overworked person in the cab doing all 3 jobs :)

@egid every Class B and Class C will require VFR departures to get a squawk code in order to effect the required radar separation standards (a discrete squawk is not, strictly speaking, necessary to do this—but it makes things much easier). The relative busyness of the airport has nothing to do with it, except insofar as it necessitates the airspace being B/C in the first place. (Whether there is a separate CD frequency or if codes are assigned on Ground Control is another question.)

@randomhead an 8 year necro, impressive! :) With the context of the previous comments, I was probably referring to the fact that B and C airports are relatively busy, compared to D or uncontrolled fields where students usually train.

This answer details the procedures used in the US. Other countries will be similar, possibly with different names for some positions.
When preparing for a flight, after their flight plan has been filed, the pilots will first talk to the clearance delivery controller. This controller clears their planned route and altitude. At larger airports this may be done via text rather than voice. The pilots may also call their company dispatch to get the route or fix any problems.
If there are significant delays in effect at the airport, the pilots may have to talk to the metering controller. This controller is responsible for holding aircraft at their gates. Weather at departure or arrival airports may reduce the number of aircraft the airport can handle. There is no sense in sending the aircraft onto the busy taxiways if they will just have to sit there and wait for weather or traffic before they can depart.  The metering controller will eventually hand the pilots off to the next controller when the aircraft is finally cleared for departure.
Again at larger airports, the next contact will be with the ramp controller. This controller is responsible for a ramp area near the terminals. The controller will let the pilots know when it's safe to push back, and which way they should turn at the end of the push back. After starting their engines, the pilots will receive instructions for exiting the ramp.
Once pilots reach the edge of the ramp area, they will call the ground controller. This controller is responsible for managing traffic on most of the taxiways. The pilots will receive instructions for taxiing to their departure runway. If this route takes them across an active runway, the ground controller will have to clear this with the tower controller, or possibly have the pilots switch to the tower controller for permission to cross the runway. At small airports, the ground controller handles all of the previous positions as well.
Upon reaching their departure runway, the pilots will switch to the tower controller. The tower controller is responsible for the active runways. Busy airports may have multiple tower controllers managing different runways. The tower will clear the pilots for takeoff, ensuring proper separation between any other arriving/departing aircraft.
Shortly after the aircraft starts climbing, the tower will hand the pilots off to the departure controller. Depending on the location and traffic levels, this could be the same controller and frequency as the approach controller. This position is part of the TRACON (Terminal Radar Approach Control) facility. High traffic areas will have their own separate TRACON facility, while in lower traffic areas this will be combined with a tower location. In other countries this is referred to as the Terminal Control Center (TCC).
The approach controller's responsibility is routing aircraft leaving airports in the area to avoid conflicts between aircraft, but often the pilots will just fly the published departure procedure. The departure controller receives reports of weather such as precipitation or icing and passes this on to other aircraft.
Upon reaching a certain altitude, the departure will hand the pilots over to the center controller. This position is part of an ARTCC (Air Route Traffic Control Center), often abbreviated to "center". The US has 22 ARTCCs, each further divided into different sectors. Sectors are consolidated between controllers depending on traffic levels. Other countries refer to this as an Area Control Center (ACC).
The center controller is responsible for traffic at higher altitudes, primarily enroute traffic. The aircraft may pass through multiple centers, and each center may have different sectors controlled by different controllers. The pilots will be handed off between these controllers. Pilots must request altitude or course changes (for efficiency or weather reasons), and controllers may issue changes for traffic reasons. Controllers also handle pilot reports of conditions such as turbulence.
If the flight is crossing large areas of ocean, the flight may be handed over to an oceanic controller. These controllers do not always have radar coverage of the aircraft they are controlling. They rely on the aircraft following their published routes and altitudes, and providing regular position reports. The lack of radar means there are also more strict rules about deviations from flight plans.
As the pilots get closer to their destination, they will start on their arrival procedure. This usually specifies altitudes and possibly speeds, so the pilots will need to plan when to start their descent in order to meet these restrictions. The center controller will know the procedure and instruct them to follow it, and possibly issue deviations as traffic requires.
At some point before arriving, the pilots may also contact their company at the arrival airport. They may need to coordinate passenger needs such as wheelchairs or medical/security issues, or communicate maintenance needs. They will also coordinate what gate they are scheduled to use. This can also be done via text, such as through ACARS.
When descending out of center's airspace, pilots will be handed over to the approach controller. This position is part of a TRACON and handles arriving traffic. Depending on the location and traffic levels, this could be the same controller and frequency as the departure controller. The pilots will continue to follow the arrival procedure, again with any deviations issued by the controllers. The departure controller receives reports of weather such as precipitation or icing and passes this on to other aircraft. The controller may also issue information about weather observed on radar, although each aircraft also has its own radar onboard. The approach controller will then provide vectors to the pilots as needed to get them to an approach procedure. The approach controller will then clear the aircraft to fly a certain approach to a certain runway.
Once the aircraft is established on the approach a few miles from the runway, the pilots will be handed over to the tower controller. The tower will clear the aircraft to land. If the aircraft has to go around, the pilots are sent back to approach to be sequenced back onto the approach. The tower controller also receives reports of wind shear and at what altitude pilots break out of the clouds, to pass on to the next aircraft.
After exiting the runway, the pilots will change to the ground controller. As before, the ground controller issues taxi instructions to get the aircraft to the appropriate ramp.
Before entering the ramp, the pilots will call the ramp controller. The ramp controller must see if the aircraft's scheduled gate is available. If not, the aircraft will have to wait on a taxiway somewhere. Once the aircraft's gate is free, the ramp controller allows the aircraft to enter the ramp, and park at the gate.

@pericynthion thanks for the suggestions, I've edited to include those things.

As others have commented, it's really not much different than the average GA communications.
Instead of communicating with an FBO, airlines communicate with their company. (The company handles the aircraft part of it, anyway.)
The clearance controller exists just as it does in GA.
There'll be a ground controller they need to ask before pushing back, but that's just because it's logistically a tad more difficult for two 737s to go around each other than two 172s.
Tower (called "local control" in ATC) is just like you're used to - control of the runways and the immediate airspace.
Jets (and turboprops...) spend less time talking to Departure/Approach, and usually end up on Center frequencies much more quickly. There's various sectors, but again... it's no different than you're used to.
The process then goes in reverse for landing as you'd expect.
All of these (even the airline company frequencies at larger airports) are available on LiveATC. Try following one flight through from start to finish sometime!

Don't forget that at the end of your flight, just like captain Sully, you also check-in with the spouse.

Air Traffic Control service is divided into three main sections:

Aerodrome Control
Approach Control
Area Control

Aerodrome Control service is provided by an aerodrome control tower. All flights at controlled aerodromes are provided with aerodrome control service.
The purpose of aerodrome control service is to separate flights in the vicinity of the aerodrome, on the runways and on the manoeuvring area.
At major aerodromes, you will find multiple different working positions in the aerodrome control tower. The three normal ones are clearance delivery, ground control and tower control. At smaller airports, these three services are all provided by the same unit.
Clearance delivery (callsign ‘Delivery’) issues pre-departure clearances to IFR traffic, usually with a clearance limit of their destination. Included in the clearance are instructions about which route to follow after departure, the cleared level and any other required information.
Ground control (callsign ‘Ground’) issues taxi instructions on the manoeuvring area. This usually does not include the apron areas where aircraft are parked (the movement area).  Some airports have separate units directing traffic on these areas, but they are not technically air traffic control units.
Tower control (callsign ‘Tower’) is responsible for traffic on the runways and in the vicinity of the aerodrome. Tower issues takeoff and landing clearances, and will direct (mainly VFR) traffic in the air close to the aerodrome.
Approach Control service can be provided by an aerodrome control tower or an area control centre, but is usually provided by an approach control unit.
The purpose of approach control service is to provide ATC to arriving and departing flights. Basically, approach control covers the part of a flight from the beginning of descend to the final approach, or from the initial climb until reaching the cruise phase.
Approach control service can be subdivided into different sectors. Their exact use vary from place to place. Common callsigns are ‘Approach’, ‘Arrival’, ‘Departure’ and ‘Precision’. As these names suggest, sometimes you have one approach control unit handling only arriving traffic while another is handling only departing traffic. You will also often find airports where one sector is responsible for the initial approach phase, while another unit is responsible for the final sequencing onto the final approach.
Area Control service is provided by an area control centre, but can also be provided by an approach control unit in control areas near large airports.
Area Control service is provided to controlled flights in control areas. This typically means the portion of flight between departure and arrival, except when covered by approach control. This includes en-route conflict solving, level changes, route changes etc. The callsign of area control units is usually ‘Control’ or ‘Radar’, except in the USA where they call it ‘Center’.
A typical IFR flight between two controlled airports will talk to the above units in the following order:

Clearance delivery, to obtain the pre-departure clearance
Ground control, to taxi from the apron to the departure runway
Tower control, to receive clearance for takeoff
Approach control (approach/departure) for the departure and climb toward cruise
Area control for the cruise portion of flight. Most flights cross through several area control sectors.
Approach control for the arrival and approach
Tower control, to receive landing clearance
Ground control, to taxi from the arrival runway to the apron

Can you award yourself your own bounty? ;)  That's my way of saying "Nice answer".


text:
Does a helicopter's boundary layer have different properties that are caused by the rotation? How does this affect stability of the helicopter?
I recently got introduced to the boundary layer theory of an airfoil, and I couldn't stop thinking of how it would be affected by rotation as well as the varying speeds of rotation.

Can you please clarify and edit the question: the question is most probably about the flow near the rotor blades not the fuselage or other surfaces, right?

You are right, the rotation does affect the boundary layer.
Normally, as a wing approaches its stall angle of attack, the boundary layer becomes thicker and the flow starts to separate near the trailing edge. On a rotating rotor or propeller blade, the slowed-down boundary layer will experience a centrifugal acceleration, so it does not come to a standstill eventually (as it does in case of separation), but merely starts to flow tipwards. Since the speed of the rotor blade increases as the flow moves towards the tip, the boundary layer experiences an additional Coriolis acceleration. Therefore, flow separation is delayed on rotors and propellers compared to the two-dimensional case (for example, when the rotor airfoil is tested in a windtunnel).
From heli-air.net:

The helicopter rotor experiments of Dwyer & McCroskey (1971) also
  suggest favorable effects on the spanwise development of the boundary
  layer, which tend to delay the onset of flow separation to a higher
  blade section AoA and thus serve to increase the maximum thrust of the
  rotor system.

The stability is affected by other effects; here the peculiarities of a boundary layer on a rotating wing make little difference beyond the higher stall angle of attack.

Rotation of the blade has a whole host of effects on the boundary layer of the blade: because of the velocity distribution, almost all aerodynamic effects are found at different radii.

If we look at the velocity distribution at forward speed, we can immediately see that the forward moving blade has a high tip speed, and the rearward moving blade has a region of reverse flow. So we can observe:

A variation in Reynolds number from very low to about 10$^6$ near the tip. Below Re = 10$^5$ the boundary layer effects are mainly laminar, above 10$^5$ mainly turbulent. So we can find separation bubbles at different chord lengths as a function of blade section radius, and of position of the blade in the rotation cycle.
A variation in Mach number from zero to critical. Shock waves may appear at the advancing blade, and disappear again later in the rotation circle. From Leishman:

If at any point during this process the shock wave becomes sufficiently strong, then the high adverse pressure gradient will cause the boundary layer to separate causing a loss of lift and an increase in drag known as shock induced stall.

The region of reverse flow is outside of centre - it still produces lift despite the reverse condition. The high tip mach may cause lift loss or rapid drag increase, resulting in roll or yaw moment.

Pitching moment of the blade also changes continuously over the rotation path. All in all the trim of the helicopter changes a lot with airspeed, and needs continuous adjustment, especially in the higher sped region. These effects are partly compensated for by blade flapping: when left free to flap like the wing of a bird, the forward blade has a tendency to rise, reducing the angle of attack, and the retreating blade falls, increasing AoA. So this flapping takes care of roll moments, but causes sine wave variations of the free stream hitting the blade. And there are many of these Unsteady Airflow effects:

Stability responses of the helicopter on these effects vary. The rotation of the blades causes a sort of averaging out of effects, and for instance angle of attack oscillations do not cause unsteady stall behaviour noticeable by the pilots: the rotor is connected to the fuselage by a hinge after all. Fuselage will follow rotor, but with a time delay that further evens out responses.
More info in Leishman, chapters 7 and 8.


text:
How can an aircraft turn if the horizontal force component is zero?
Here is the simplest coordinated turn diagram

Picture from https://physics.stackexchange.com/questions/100169/pendulum-in-a-turning-airplane
Being:
$L$ = aerodynamic lift
$\tau$ = angle between vertical component and lift
$Fc$ = centrifugal "force"
In many books you can read that:
$L\cdot sin(\tau) = Fc = m \cdot (V^2)/R$
 where $R$ is the curvature radius.
Then, how can an aircraft turn if there is no net force? I mean, from Newton's second law we have that $F = m \cdot a$ and in this case centrifugal force compensates for lift component in the horizontal plane.

You skipped a step. The image depicts forces acting on a pendulum, not just the plane. You can't just take random images from the internet and expect them to make sense in a different context :) The two forces are weight and lift, and they combine to make an apparent force on the pendulum. The horizontal component of the lift produces an apparent force on the contents of the airplane, which changes the deviation of a suspended object from the weight vector (in this case, by exactly the same amount as the "floor" of the aircraft, which is the point of the original question).

That diagram doesn't what is applying which forces to what,

A duplicate question has now been asked and has received several answers -- https://aviation.stackexchange.com/q/101394/34686

The problem of the picture you are looking at is that both the actual and the apparent forces are shown.
The real force is the centripetal one (that in turn is only the horizontal component of the lift force), the centrifugal one does not exists, is only an "impression" of the centripetal one as seen by a person standing in the aircraft (a non inertial reference frame).
The aircraft flies in a circle because the centripetal component pulls the aircraft towards the center of the turn, the centrifugal force is only what a person in the aircraft feels, but is not an actual force acting on the aircraft, it is a representation of the inertia of the aircraft (and person inside).

so, a personal crew will sense that "forcé", right?

@XF-91 yes, any person inside the aircraft

thank you very much. A last question: so, in an inertial frame of reference (Earth in first aproximation) the only forces that exist are lift, weight, and centripetal forcé, right. Thank you again

@XF-91 the centripetal force (no accent) is a component of lift, not an independent force, I'll improve the answer.

@XF-91 the only forces are aerodynamic forces (usually decomposed as lift and drag) and weight. Nothing else act on the airplane, and thus no other forces exist.

Mandatory XKCD: xkcd.com/123

A frame in which the centrifugal force exists (as mentioned in the XKCD comic) must be rotating relative to the inertial frame. That is, the only way to legitimately assume a centrifugal force is if you have already assumed that the aircraft is turning.

I know that schools are teaching apparent forces, but the naming is misleading and better be avoided. Again, see Newton's 3rd.

@Jenc In any case you cannot write the the centrifugal force and the centripetal forces as if acting on the same object because they are not. The centrifugal force acts on the plane and the centripetal force on the air.  If both acted on the plane, the plane would indeed be in balance and not turn.

@MichaelKarnerfors I am glad that we agree on all these points.

Alternatively, when you do the math in the reference frame of the plane, the centrifugal force is exactly the one that prevents the plane from turning (since the plane can't turn in its own frame). To answer the question: the plane indeed does not turn if you consider the centrifugal force, because it means you have chosen to use its own reference frame.

I was wrong, above comment withdrawn. Centripetal and centrifugal forces cannot co-exist in the same reference frame.

Since a passenger feels both the centrifugal force and the weight when the airplane is making the turn, does the normal force exerted by the airplane on it increase in the turn compared to when the flight is horizontal?

@XF-91 yes, they would feel heavier. Look up "g factor" or "load factor"

the centrifugal force is not an impression of the centripetal force, but of the rotating frame of reference. The centripetal force only allows the rotating frame of reference to be relevant by keeping the airplane stationary in it.

@Federico: Summarizing to close the question -> In an analysis using an inertial frame of reference, in the diagram of forces we would draw the lift and the weight, whereas for a passenger located inside the airplane we would only show its weight and the centrifugal force. This is so?

@XF-91 almost. For the passenger you still have to consider the force that the seat is imparting, that is roughly equivalent to the lift.

@Federico: but if we consider the aerodynamic lift besides the weight and centrifugal force we have the same situation that at the beginning, there's no net force acting on the passenger (weightlessness situation). Is it not precisely the centrifugal force that produces this pushing effect of the seats on the passenger?

@XF-91 the passenger is not going anywhere w.r.t. the airplane, right? or is flying away from the seat?

@Federico: Sorry, I was forgetting the normal force exerted by the seat on the passenger (in this case it should not include the lift, right?). I'm making a mess :)

@XF-91 and I said "roughly equivalent to the lift" I didn't say "that is the lift". yes, you should go to chat.

The problem is not that the picture mixes “real” and “apparent” forces, the problem is that it shows the forces as they exist in the reference frame of the aircraft, where it indeed does not turn. The weight is also and inertial (i.e. “apparent”) force; in Newtonian mechanics partly (it is a sum of gravity and centrifugal forces due to rotation and orbiting of Earth), in general relativity completely (because in general relativity, reference frames in free fall take the special role of inertial reference frames and gravity becomes an inertial force too).

It is easier if we look only at the forces experienced by the aircraft, and in an inertial frame of reference

In this revised diagram, the vertical component of the lift balances the weight, which is vertical. There is a remaining horizontal component of the lift, and this causes the turning.
"Centrifugal force" does not exist (and is not needed) in an inertial frame of reference
The problem with the original diagram in the question is it superimposes an imaginary force, the centrifugal force, upon the real list of forces on the aircraft.
It is hard for the general public to understand Newton's first law of motion, that any object tends to travel in a straight line when no force is acting it. It is hard for them to understand that motion in a circle is dramatically different from straight-line, constant-speed motion, since both seem (in a sense) steady or continuous.
"Centrifugal force" is a term produced by humans to describe what they think must be happening
In the case of a passenger going round a corner in the car, this inward force is applied by the outer door of the car, against which the passenger ends up leaning during the turn. It is an inward (centripetal) force, but for the general public who is not aware of the special status of inertial frames of reference will likely use the car (which seems large and solid and therefore plausible) as a frame of reference and say "something must be pushing me against the door" rather than "the door is pushing me towards the centre of the circle".
Because of this use of a non-inertial frame of reference, the general public explanation needs to an extra force to explain why objects seem to accelerate towards the outer door of the car and the passenger feels pressed against it. In an inertial frame of reference it is the outer door of the car pressing in on the passenger, and accelerating towards any objects moving freely inside the car.
In everyday life we sometimes use special frames of reference for convenience
We say the sun "comes up" in the morning, rather than the earth "rotates bringing the sun into view". This is because it is convenient for humans to use a non-inertial, but readily identifiable, frame of reference. It does not mean that people really think that the sun orbits around a stationary earth.
It is perfectly reasonable to imagine the sun travelling around the earth, for convenience, but if we want to make calculations about space travel with multiple objects (stars, planets etc) affecting a body's motion, it will get very complicated indeed to add in all the numerous centrifugal forces etc we would have to add in. It would be simpler to use an inertial frame of reference, and all objects moving relatively simply in relation to that.
Likewise we can learn to be sophisticated about circular motion. Let the general public speak of "centrifugal force" as this simplifies conversation, but when trying to calculate things, use an inertial frame of reference to avoid agony.
Even apparently inertial frames of reference may have unconsidered rotation  in relation to a larger-scale environment
Jan Hudec in the comments below brilliantly deflates the above criticism of rotating frames of reference by pointing out that even my notionally inertial frame of reference (the earth) is in fact rotating, so that the gravitational force measured is lower than it would be for a plane moving similarly above a non-rotating earth, i.e. is attenuated by a centrifugal component. Ouch!
And of course the earth is rotating around the sun, and the sun around the galaxy, etc:
Big fleas have little fleas,
Upon their backs to bite 'em,
And little fleas have lesser fleas,
and so, ad infinitum.

So it is just a question of how much rotation one wants to accept as part of one's frame of reference. My suggestion is that when a questioner is feeling puzzled as the original poster was, a good way to break the problem down is to keep the frame of reference outside the rotation being studied, so that the studied rotation appears as a rotation: in this case the only acceleration (and therefore force on the orbiting object) will be centripetal.

This has already been mentioned in a comment on another answer, but here it is again: https://xkcd.com/123/  Randall Munroe appears to have studied physics beyond the level of a first-year college course.

@DavidK to be noted that in the comic it is correctly written "you will feel centrifugal force". The central concept is feeling.

"It does not mean that people really think that the sun orbits around a stationary earth." - Clearly you have not spent much time reading YouTube comments, hehe.

@Federico The physics concept is to construct Newton's laws in a rotating frame of reference, as described in the second and third frames of the comic. That's not merely subjective feeling, but the mathematics is beyond what you will get in the usual popular science treatment.

I do think this is the best answer, because of the diagram, but I'm reluctant to upvote something that says "centrifugal force is a mistaken [...] concept" unconditionally (especially in large bold text). The xkcd comic people have been linking has it right: it's a perfectly valid concept, just one that only exists in noninertial reference frames.

" Technically invalid reference frames" is equally misleading. Any reference frame is equally valid, technically and mathematically.  Some are a bit more convenient, but that is engineering and not science.

@David Z, David K, Federico and MSalters (and Jason C!), Sorry yes you are right, I have revised wording to be more diplomatic. Its just that my brain isn't big enough to cope with non-inertial reference frames because all the little fiddly things you have to do, not only centrifugal forces (and keeping track of what forces act on what) but also Coriolis forces etc. My head explodes 8-)

I suppose that's good enough for an upvote, then. I do still think this answer is a little more dismissive of noninertial reference frames than it should be, but it's not inaccurate anymore in the sense that I thought it was before.

Note that in general relativity, reference frames in free fall take the special place of the inertial reference frames of Newtonian mechanics and general relativity and gravity becomes just another inertial force.

Also note that weight is a sum of gravity and centrifugal force due to rotation of the Earth. Because reference frame of Earth is not inertial either.

@Jan Hudec - Brilliant! I had completely missed this. Am adding in to the answer...

The diagram is not a correct force diagram in the inertial frame of reference.
In the inertial frame of reference, the horizontal force on the aircraft is unbalanced. This causes the aircraft to accelerate in the direction of the net horizontal force, perpendicular to the direction of flight, and this curves the path of the aircraft.
The law of "equal and opposite reaction" does not mean that every force on every object is balanced at all times. If the forces were always balanced, nothing would ever move. The aircraft could not leave its parking space, let alone get airborne and turn.
Physicists sometimes consider non-inertial frames in which there actually is a centrifugal force on the aircraft as shown in the figure.
In order to construct such a frame of reference, however, the physicist assumes that the frame of reference is rotating relative to an inertial frame of reference. During a turn, the frame of reference attached to the aircraft is such a frame of reference, but only because the aircraft is turning.
In other words, the forces shown in the figure cannot all exist in an inertial frame of reference, and cannot exist in the frame of reference of an aircraft that is flying straight (without turning), but they do all exist in the non-inertial frame of reference of an aircraft that is turning.
No matter how you look at it, a horizontal component of lift by banking the aircraft must be associated with a turn.

Lift and weight are acting on the plane. If the vector sum of lift and plane is not zero, the plane will accelerate.
When the plane is banked:

the lift is not vertical
the lift is not parallel to the weight
the sum of lift and weight is not zero
the sum of lift and weight has a horizontal component (centripetal force)
the centripetal force causes the plane to turn

The concept of centrifugal force exists only in a rotating reference frame. Therefore the drawing with centripetal and centrifugal forces acting in the same reference frame is wrong.

the aircraft is not a closed system.

No, but air space is (as far as the relevant forces are concerned).

I have issues with your definition of "closed system" then

You may name them.

a closed system is a system that cannot interact with anything outside of it. air interacts with everything around it.

It appears that your closed system is too small. Allow it to be large enough.

Let us continue this discussion in chat.

The simplest and best answer to the question:  An aircraft can't!
Well......with a few exceptions:
Blimps and airships turn using a rudder input just like their nautical cousins do.
Airplanes can be turned using only rudder inputs as well, thereby creating a centripetal component of thrust.  This type of turning is far from efficient though and not practical for conventional maneuvering.
Also remember that centrifugal force is a reaction experienced by the aircraft and its occupants to the centripetal loading from the horizontal component of lift.  There is no external centrifugal force being applied to the airplane pulling it against the turn as in the diagram.

I think what you're saying is that an airplane cannot practically turn without banking. But I think OP is asking about a banked turn, claiming that the force diagram indicates that that won't work either.